1 + 2w +3w2 +..... nwn-1
find the sum where,w = nth root of unity
11Does this have something to do with differentiation? Cos i notice that it is f'(ω) where f(x)=c+x+x2+x3+........+xn.[7]
62Aragon...
going by ur method
f(x)=1+x+x2+x3+........+xn
f(x) = (1-xn+1)/(1-x)
f'(x) = -(n+1)xn/(1-x) + ((1-xn+1)/(1-x)2)
substitute x= ω
You will get the answer :)
62something may be wrong... in the above thing... if it is do correct it :)
btw aragon it is also a simple Arithemetic Geometric Series...
see if u are able to see that :!)
11Yes. I'm noticing that now. but at first sight, the only thing that came to me was differentiation. But I didn't know the alternative expression for f(x).
S= 1+2ω+3ω2+...............nωn-1
ωS= ω+2ω2+............+n-1ωn-1+nωn
S-Sω= 1+ω+ω2+...........+ωn-1-nωn
S=(1-ωn)/(1-ω)2 - nωn/(1-ω)
=(1-ωn-nωn+nωn+1)/(1-ω)2
=[1-(n+1)ωn+nωn+1]/(1-ω)2
Is this anywhere near the answer? [2]
1nishant,aragorn, thanks for the solution
@ Aragorn,you got the answer
S-Sω= 1+ω+ω2+...........+ωn-1-nωn
=-nωn (as 1+ω+ω2+...........+ωn-1=0)
S= n/(ω-1)
11Can anyone pls give me the proof for 1+ω+ω2+.........+ωn-1=0?
62aragon the proof is simple..
this is the sum of all the roots of
1-xn=0
Sum of the roots of a polynomial is? : (in terms of coefficients?)
11Thanx Vivek. [1]
Bhaiyyah, understood vivek's proof but didn't get ur hint. Sorry [2]
11But here co-eff of -xn-1 is 0. So sum is 0? Thanx machan [1]
